AGC015 C-Nuske vs Phantom Thnook AtCoder 思路 前缀和

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目录

• [TOC]

题目链接

AGC015 C-Nuske vs Phantom Thnook AtCoder

题解

树的性质有:
如果每个蓝色连通块都是树，那么连通块个数=总点数−总边数。
二维前缀和维护点数和边数。
(O(nm + q))

代码

#include <cstdio>
#include <iostream>
#include <algorithm> 
#define gc getchar() 
#define pc putchar
#define LL long long
inline int read() { 
	int x = 0,f = 1; 
	char c = getchar(); 
	while(c < '0' || c > '9') c = gc; 
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar(); 
	return x * f; 
}
void print(int x) { 
	if(x < 0) { 
		pc('-'); 
		x = -x; 
	} 
	if(x >= 10) print(x / 10); 
	pc(x % 10 + '0'); 
} 
const int maxn = 2010; 

int n,m,q;
int a[maxn][maxn];
int b[maxn][maxn],c[maxn][maxn];
char s[2007]; 
inline int calc(int a[maxn][maxn],int x1,int y1,int x2,int y2) { 
    if(x1 > x2 || y1 > y2)  return 0; 
    return a[x2][y2] - a[x1 - 1][y2] - a[x2][y1 - 1] + a[x1 - 1][y1 - 1]; 
} 
int main() { 
	n = read(),m = read(),q = read(); 
    for(int i = 1;i <= n;++ i) { 
        scanf("%s",s + 1); 
		for(int j = 1;j <= m;++ j) 
            a[i][j] = s[j] - '0'; 
    }  
    for(int i = 2;i <= n;++ i) 
        for(int j = 1;j <= m;++ j) 
            b[i][j] = a[i][j] & a[i - 1][j]; 
    for(int i = 1;i <= n;++ i) 
        for(int j = 2;j <= m;++ j) 
            c[i][j] = a[i][j] & a[i][j - 1]; 
    for(int i = 1;i <= n;++ i) 
        for(int j = 1;j <= m;++ j) { 
            a[i][j] = a[i][j] + a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];  
            b[i][j] = b[i][j] + b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1]; 
            c[i][j] = c[i][j] + c[i - 1][j] + c[i][j - 1] - c[i - 1][j - 1]; 
    } 
    while(q -- ) { 
        int x1 = read(),y1 = read(),x2 = read(),y2 = read(); 
        print(calc(a,x1,y1,x2,y2) - calc(b,x1 + 1,y1,x2,y2) - calc(c,x1,y1 + 1,x2,y2)); 
        pc('
'); 
    } 
    return 0; 
}